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花花酱 LeetCode 2227. Encrypt and Decrypt Strings

You are given a character array keys containing unique characters and a string array values containing strings of length 2. You are also given another string array dictionary that contains all permitted original strings after decryption. You should implement a data structure that can encrypt or decrypt a 0-indexed string.

A string is encrypted with the following process:

  1. For each character c in the string, we find the index i satisfying keys[i] == c in keys.
  2. Replace c with values[i] in the string.

A string is decrypted with the following process:

  1. For each substring s of length 2 occurring at an even index in the string, we find an i such that values[i] == s. If there are multiple valid i, we choose any one of them. This means a string could have multiple possible strings it can decrypt to.
  2. Replace s with keys[i] in the string.

Implement the Encrypter class:

  • Encrypter(char[] keys, String[] values, String[] dictionary) Initializes the Encrypter class with keys, values, and dictionary.
  • String encrypt(String word1) Encrypts word1 with the encryption process described above and returns the encrypted string.
  • int decrypt(String word2) Returns the number of possible strings word2 could decrypt to that also appear in dictionary.

Example 1:

Input
["Encrypter", "encrypt", "decrypt"]
[[['a', 'b', 'c', 'd'], ["ei", "zf", "ei", "am"], ["abcd", "acbd", "adbc", "badc", "dacb", "cadb", "cbda", "abad"]], ["abcd"], ["eizfeiam"]]
Output

[null, “eizfeiam”, 2]

Explanation Encrypter encrypter = new Encrypter([[‘a’, ‘b’, ‘c’, ‘d’], [“ei”, “zf”, “ei”, “am”], [“abcd”, “acbd”, “adbc”, “badc”, “dacb”, “cadb”, “cbda”, “abad”]); encrypter.encrypt(“abcd”); // return “eizfeiam”.   // ‘a’ maps to “ei”, ‘b’ maps to “zf”, ‘c’ maps to “ei”, and ‘d’ maps to “am”. encrypter.decrypt(“eizfeiam”); // return 2. // “ei” can map to ‘a’ or ‘c’, “zf” maps to ‘b’, and “am” maps to ‘d’. // Thus, the possible strings after decryption are “abad”, “cbad”, “abcd”, and “cbcd”. // 2 of those strings, “abad” and “abcd”, appear in dictionary, so the answer is 2.

Constraints:

  • 1 <= keys.length == values.length <= 26
  • values[i].length == 2
  • 1 <= dictionary.length <= 100
  • 1 <= dictionary[i].length <= 100
  • All keys[i] and dictionary[i] are unique.
  • 1 <= word1.length <= 2000
  • 1 <= word2.length <= 200
  • All word1[i] appear in keys.
  • word2.length is even.
  • keysvalues[i]dictionary[i]word1, and word2 only contain lowercase English letters.
  • At most 200 calls will be made to encrypt and decrypt in total.

Solution:

For encryption, follow the instruction. Time complexity: O(len(word)) = O(2000)
For decryption, try all words in the dictionary and encrypt them and compare the encrypted string with the word to decrypt. Time complexity: O(sum(len(word_in_dict))) = O(100*100)

Worst case: 200 calls to decryption, T = 200 * O(100 * 100) = O(2*106)

C++

Optimization

Pre-compute answer for all the words in dictionary.

decrypt: Time complexity: O(1)

C++

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